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Math Help - Binomial Expansion

  1. #1
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    Binomial Expansion

    Expand the first 3 terms in [6-13x+6x^2]^5
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Punch View Post
    Expand the first 3 terms in [6-13x+6x^2]^5
    when using expansion of (a+b)^n

     T_{k+1}=\displaystyle \binom {n}{k} a^{n-k} \cdot b^{k}

    that's to find any member (first, second, ... )

    so u try and see where do u go with this
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  3. #3
    MHF Contributor Unknown008's Avatar
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    What you can do is treat it like this:

    [6 - (13x - 6x^2)]^5

    Then apply the formula to find the first three terms.

    (a + b)^n = a^n + \binom{n}{1} (a)^{n-1}(b) + \binom{n}{2} (a)^{n-2}(b)^2+...

    Taking a = 6
    b = - (13x - 6x^2)
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  4. #4
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    that makes sense thanks
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  5. #5
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Unknown008 View Post
    What you can do is treat it like this:

    [6 - (13x - 6x^2)]^5

    Then apply the formula to find the first three terms.

    (a + b)^n = a^n + \binom{n}{1} (a)^{n-1}(b) + \binom{n}{2} (a)^{n-2}(b)^2+...

    Taking a = 6
    b = - (13x - 6x^2)
    perhaps that's the better way to go

    formula that i posted is the same, but way that it's wrote it's way easier to find let's say 181 element of some expansion not for this problem but works the same
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  6. #6
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    Quote Originally Posted by Punch View Post
    Expand the first 3 terms in [6-13x+6x^2]^5
    I have a question about the question itself.
    Who is to say which are the first three terms?
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by Plato View Post
    I have a question about the question itself.
    Who is to say which are the first three terms?
    Yes, the question should have included something like: "the first three terms in ascending powers of x"
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  8. #8
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Plato View Post
    I have a question about the question itself.
    Who is to say which are the first three terms?
    I was wondering that too, figured the only natural choices were: (1) constant term, x^1 term, x^2 term; (2) x^10 term, x^9 term, x^8 term; and that since the trinomial we are expanding is given in form (1) it would be acceptable to assume form (1).
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  9. #9
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    Quote Originally Posted by Punch View Post
    Expand the first 3 terms in [6-13x+6x^2]^5
    You can also expand

    \left(x[6x-13]+6\right)^5

    It may be more compact.
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  10. #10
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    Hello, Punch!

    This is a messy one . . . even with a good procedure.


    \text{Find the first 3 terms of: }\;(6-13x+6x^2)^5

    It is evident that the first three terms are:
    . . . . \text{constant},\;x\text{-term},\;x^2\text{-term}

    \text{We have: }\:(a + b + c)^5\quad \text{where: }\:\begin{Bmatrix}a &=& 6 \\ b &=& \text{-}13x \\ c &=& 6x^2 \end{Bmatrix}


    The "Trinomial Expansion" forms an array:

    \begin{array}{cccccccccccc}&&& {5\choose5,0,0}a^5 \\ && {5\choose4,1,0}a^4b && {5\choose4,0,1}a^4c \\ & {5\choose3,2,0}a^3b^2 && {5\choose3,1,1}a^3bc && {5\choose3,0,2}a^3c^2 \\ {5\choose2,3,0}a^2b^3 && {5\choose2,2,1}a^2b^2c && {5\choose2,1,2}a^2bc^2 && {5\choose2,0,3}a^2c^3 \\ \vdots && \vdots && \vdots && \vdots\end{array}



    We want only the first four terms (reading from the top).
    . . [The rest contain x^3 or higher powers.]


    . . {5\choose5,0,0}(6^5) + {5\choose4,1,0}(6^4)(\text{-}13x) + {5\choose4,0,1}(6^4)(6x^2) + {5\choose3,2,0}(6^3)(\text{-}13x)^2

    . . =\;1(7776) + 5(1296)(\text{-}13x) + 5(1296)(6x^2) + 10(216)(169x^2)

    . . =\;7,\!776 - 84,\!240x + 38,\!880x^2 + 365,\!040x^2

    . . =\;7,\!776 - 84,\!240x + 403,\!920x^2



    But check my work . . . please!
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  11. #11
    Senior Member yeKciM's Avatar
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    or look at it like this

    (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5


    and so when doing u'll get

    T_1= \displastyle \binom {5}{0} (-x)^5 [(x-1)(6x-6)]^0=-x^5

    T_2= \displastyle \binom {5}{1} (-x)^4 [(x-1)(6x-6)]^1=5x^4[6x^2-12x+6]=30x^6-60x^5+30x^4

    T_3= \displastyle \binom {5}{2} (-x)^3 [(x-1)(6x-6)]^2=-10x^3[(x-1)^2(6x-6)^2]=-360x^7+1440x^6-2160x^5+1440x^4-360x^3

    u can go further... but u get the idea

    or am i wrong here ?
    Last edited by yeKciM; August 6th 2010 at 07:36 AM.
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  12. #12
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Soroban View Post
    . . =\;7,\!776 - 84,\!240x + 403,\!920x^2


    But check my work . . . please!
    Confirmed with Mathematica. Unknown008's method gives the same result.
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  13. #13
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by yeKciM View Post
    or look at it like this

    (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5


    and so when doing u'll get

    T_1= \displastyle \binom {5}{0} (-x)^5 [(x-1)(6x-6)]^0=-x^5

    T_2= \displastyle \binom {5}{1} (-x)^4 [(x-1)(6x-6)]^1=5x^4[6x^2-12x+6]=30x^6-60x^5+30x^4

    T_3= \displastyle \binom {5}{2} (-x)^3 [(x-1)(6x-6)]^2=-10x^3[(x-1)^2(6x-6)^2]
    =-360x^7+1440x^6-2160x^5+1440x^4-360x^3

    u can go further... but u get the idea

    or am i wrong here ?
    This makes it harder to find the first three terms, compared with what Unknown008 posted, doesn't it?
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  14. #14
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by undefined View Post
    This makes it harder to find the first three terms, compared with what Unknown008 posted, doesn't it?



    hmmmm

    probably lost in translation, but if u look at (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5 then first one should be :

     T_1=\displaystyle \binom{5}{0} a^{5-0}\cdot b^0

    and so on...

    but probably i'm little (or a lot) of with meaning of "terms" for me it's like n-th member or something like that


    Edit: Sorry, but now i'm little a bit of confused
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  15. #15
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by yeKciM View Post
    hmmmm

    probably lost in translation, but if u look at (6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5 then first one should be :

     T_1=\displaystyle \binom{5}{0} a^{5-0}\cdot b^0

    and so on...

    but probably i'm little (or a lot) of with meaning of "terms" for me it's like n-th member or something like that


    Edit: Sorry, but now i'm little a bit of confused
    There are 11 terms in the expansion of (6-13x+6x^2)^5, starting with the constant term and ending with the x^{10} term. Soroban gave the first three terms in post #10.
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